The volume of the 0.279 M Ca(OH)₂ solution required to neutralize 24.5 mL of 0.390 M H₃PO₄ is 51.4 mL
<h3>Balanced equation </h3>
2H₃PO₄ + 3Ca(OH)₂ —> Ca₃(PO₄)₂ + 6H₂O
From the balanced equation above,
- The mole ratio of the acid, H₃PO₄ (nA) = 2
- The mole ratio of the base, Ca(OH)₂ (nB) = 3
<h3>How to determine the volume of Ca(OH)₂ </h3>
- Molarity of acid, H₃PO₄ (Ma) = 0.390 M
- Volume of acid, H₃PO₄ (Va) = 24.5 mL
- Molarity of base, Ca(OH)₂ (Mb) = 0.279 M
- Volume of base, Ca(OH)₂ (Vb) =?
MaVa / MbVb = nA / nB
(0.39 × 24.5) / (0.279 × Vb) = 2/3
9.555 / (0.279 × Vb) = 2/3
Cross multiply
2 × 0.279 × Vb = 9.555 × 3
0.558 × Vb = 28.665
Divide both side by 0.558
Vb = 28.665 / 0.558
Vb = 51.4 mL
Thus, the volume of the Ca(OH)₂ solution needed is 51.4 mL
Learn more about titration:
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I think the answer is A but i'm not sure
Answer:
B 1.23 g/cc
Explanation:
For something to float on seawater, the density must be less than 1.03 g/mL. If the object sinks, the density is greater than 1.03 g/mL.
Let’s examine the answer choices. Keep in mind, the ice berg is mostly below the water level.
A. 0.88 g/cc
This is less than 1.03 g/cc, which would result in floating.
B. 1.23 g/cc
This is the best answer choice. The iceberg is mostly beneath the water, but some of it is exposed. The density is greater than 1.03 g/mL, but not so much greater that it would immediately sink.
C. 0.23 g/cc
This is less than 1.03 g/cc, which would produce floating.
D. 4.14 g/cc
This is much greater than 1.03 g/cc and the result would be sinking.
Answer: The given pairs of electrons most likely reside in 
type of orbital.
Explanation:
As it is given that two lone pair of electrons are present on the oxygen atom of ketone (such as cyclohexanone).
Also, there will be one bond pair between carbon and oxygen atom.
Hence, total electrons present in the domain are as follows.
2 lone pairs + 1 bon pair of electron = 3 electron domains
This means that there will be type of orbital present.
Thus, we can conclude that given pairs of electrons most likely reside in type of orbital.
